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Chapter 6: Q4E (page 358)

Let X be a random variable for which \({\bf{E}}\left( {\bf{X}} \right){\bf{ = \mu }}\)and\({\bf{Var}}\left( {\bf{X}} \right){\bf{ = }}{{\bf{\sigma }}^{\bf{2}}}\).Construct a probability distribution for X such that \({\bf{P}}\left( {\left| {{\bf{X - \mu }}} \right| \ge {\bf{3\sigma }}} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{9}}}\)

Short Answer

Expert verified

X

P(X=x)

\(\mu - 3\sigma \)

\(\frac{1}{{18}}\)

\(\mu \)

\(\frac{8}{9}\)

\(\mu + 3\sigma \)

\(\frac{1}{{18}}\)

Total

1

Step by step solution

01

Given information

We need to construct a probability distribution for a random variable X such that \(P\left( {\left| {X - \mu } \right| \ge 3\sigma } \right) = \frac{1}{9}\) .

02

Step-2: Construction of probability distribution

Let us assume the random variable X assumes three values \(\mu - 3\sigma ,\mu ,\mu + 3\sigma \)

It is given that \(P\left( {\left| {X - \mu } \right| \ge 3\sigma } \right) = \frac{1}{9}\) .

\( \Rightarrow P\left( {X = \mu - 3\sigma } \right) = P\left( {X = \mu + 3\sigma } \right) = \frac{1}{{18}}\)

Hence \(\begin{aligned}{}P\left( {X = \mu } \right) &= 1 - \left( {\frac{1}{{18}} + \frac{1}{{18}}} \right)\\ &= \frac{8}{9}\end{aligned}\)

Hence the probability distribution

X

P(X=x)

\(\mu - 3\sigma \)

\(\frac{1}{{18}}\)

\(\mu \)

\(\frac{8}{9}\)

\(\mu + 3\sigma \)

\(\frac{1}{{18}}\)

Total

1

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Most popular questions from this chapter

Suppose that 75 percent of the people in a certain metropolitan area live in the city and 25 percent of the people live in the suburbs. If 1200 people attending a certain concert represent a random sample from the metropolitan area, what is the probability that the number of people from the suburbs attending the concert will be fewer than 270?

Suppose that 30 percent of the items in a large manufactured lot are of poor quality. Suppose also that a random sample of n items is to be taken from the lot, and let \({Q_n}\) denote the proportion of the items in the sample that are of poor quality. Find a value of n such that Pr(0.2 鈮� \({Q_n}\)鈮� 0.4) 鈮� 0.75 by using

(a) the Chebyshev inequality and

(b) the tables of the binomial distribution at the end of this book.

Let X have the binomial distribution with parameters n and p. Let Y have the binomial distribution with parameters n and p/k with k > 1. Let \(Z = kY\).

a. Show that X and Z have the same mean.

b. Find the variances of X and Z. Show that, if p is small, then the variance of Z is approximately k times as large as the variance of X.

c. Show why the results above explain the higher variability in the bar heights in Fig. 6.2 compared to Fig. 6.1.

Suppose that 16 digits are chosen at random with replacement from the set {0,...,9}. What is the probability that their average will lie between 4 and 6?

Suppose that the distribution of the number of defects on any given bolt of cloth is the Poisson distribution with a mean 5, and the number of defects on each bolt is counted for a random sample of 125 bolts. Determine the probability that the average number of defects per bolt in the sample will be less than 5.5.
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